L'Hôpital's Rule

Theorem Statement

Basic Form

where $f’(x)$ and $g’(x)$ are the derivatives of $f(x)$ and $g(x)$ respectively.

Extended Form

This rule also applies to one-sided limits and limits as $x$ approaches infinity or negative infinity.

Conditions for Application

L’Hôpital’s Rule can be applied when:

1. The limit is in an indeterminate form: $\frac{0}{0}$ or $\frac{\infty}{\infty}$
2. The functions $f(x)$ and $g(x)$ are differentiable near $x = a$ (except possibly at $a$)
3. $g’(x) ≠ 0$ near $x = a$ (except possibly at $a$)

Other Indeterminate Forms

L’Hôpital’s Rule can be adapted to handle other indeterminate forms by first transforming them:

1. 0 · ∞: Rewrite as $\frac{0}{\frac{1}{\infty}}$ or $\frac{\infty}{\frac{1}{0}}$
2. ∞ - ∞: Find a common denominator or use algebraic manipulation
3. 0⁰, 1^∞, ∞⁰: Take the natural logarithm first, then apply L’Hôpital’s Rule

Repeated Application

If after applying L’Hôpital’s Rule once, the resulting limit is still indeterminate, the rule can be applied repeatedly until a determinate form is reached.

Practical Examples

Example 1: Basic Application

Both numerator and denominator approach 0 as $x \to 0$, giving the indeterminate form $\frac{0}{0}$.

Apply L’Hôpital’s Rule: \(\lim_{x \to 0} \frac{\sin x - x}{x^3} = \lim_{x \to 0} \frac{\cos x - 1}{3x^2}\)

This is still $\frac{0}{0}$, so apply L’Hôpital’s Rule again: \(\lim_{x \to 0} \frac{\cos x - 1}{3x^2} = \lim_{x \to 0} \frac{-\sin x}{6x}\)

Apply L’Hôpital’s Rule a third time: \(\lim_{x \to 0} \frac{-\sin x}{6x} = \lim_{x \to 0} \frac{-\cos x}{6} = \frac{-1}{6}\)

Therefore, $\lim_{x \to 0} \frac{\sin x - x}{x^3} = -\frac{1}{6}$

Example 2: Limit at Infinity

As $x \to +\infty$, both numerator and denominator approach infinity, giving the indeterminate form $\frac{\infty}{\infty}$.

Apply L’Hôpital’s Rule: \(\lim_{x \to +\infty} \frac{x - \cos x}{x + \sin^2 x} = \lim_{x \to +\infty} \frac{1 + \sin x}{1 + 2\sin x \cos x}\)

As $x \to +\infty$, this approaches $\frac{1 + \sin x}{1 + \sin(2x)}$, which oscillates. However, since $\sin x$ and $\sin(2x)$ are bounded, we can analyze the dominant terms:

Therefore, $\lim_{x \to +\infty} \frac{x - \cos x}{x + \sin^2 x} = 1$

Example 3: Taylor Series Alternative

This is of the form $\frac{0}{0}$. We could apply L’Hôpital’s Rule repeatedly, but using Taylor series is more efficient:

The Taylor series for $\cos x$ is: \(\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\)

Substituting: \(\frac{\cos x - 1 + \frac{x^2}{2}}{x^4} = \frac{1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots - 1 + \frac{x^2}{2}}{x^4} = \frac{\frac{x^4}{24} - \cdots}{x^4} = \frac{1}{24} - \cdots\)

Therefore, $\lim_{x \to 0} \frac{\cos x - 1 + \frac{x^2}{2}}{x^4} = \frac{1}{24}$

Common Pitfalls and Limitations

1. Not checking for indeterminate forms: L’Hôpital’s Rule applies only to indeterminate forms, not to all limits involving fractions.

2. Incorrect differentiation: Careful differentiation of both numerator and denominator is essential.

3. Not recognizing when to stop: If after applying L’Hôpital’s Rule, the limit is no longer indeterminate, additional applications are unnecessary and may lead to errors.

4. Algebraic simplification first: Sometimes algebraic simplification can avoid the need for L’Hôpital’s Rule entirely.

Comparison with Other Methods

Taylor Series Method

For limits as $x \to 0$, Taylor series expansion often provides a more direct approach, especially for functions with well-known expansions.

Algebraic Manipulation

Simple algebraic techniques like factoring, rationalizing, or finding common denominators may be more efficient for certain limits.

Substitution Method

For some limits, especially those involving composed functions, a suitable substitution can simplify the problem significantly.

Exercises

Exercise 1: Basic Application

Solution:

1. We have the indeterminate form $\frac{0}{0}$
2. Apply L’Hôpital’s Rule: $\lim_{x \to 0} \frac{\cos x - 1}{3x^2}$
3. Still indeterminate, apply again: $\lim_{x \to 0} \frac{-\sin x}{6x}$
4. Still indeterminate, apply once more: $\lim_{x \to 0} \frac{-\cos x}{6} = -\frac{1}{6}$

Exercise 2: Limit at Infinity

Solution:

1. We have the indeterminate form $\frac{\infty}{\infty}$
2. Apply L’Hôpital’s Rule: $\lim_{x \to +\infty} \frac{1 + \sin x}{1 + 2\sin x \cos x}$
3. As $x \to +\infty$, the oscillating terms are bounded, so the limit equals 1

Exercise 3: Using Taylor Series

Solution:

1. Taylor expansion of $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$
2. Substituting: $\lim_{x \to 0} \frac{\frac{x^4}{24} - \cdots}{x^4} = \frac{1}{24}$

See:

• Limits
• Taylor Series

• Derivatives

  Exercises

1. Compute the following limits both applying De L’Hôpital rule and the Taylor formula (when possible).

a) $\lim_{x \to 0} \frac{\sin x - x}{x^3}$ b) $\lim_{x \to +\infty} \frac{x-\cos x}{x+\sin^2 x}$ c) $\lim_{x \to 0} \frac{\cos x - 1 - x^2}{x^4}$

1. Compute the following limits both applying De L’Hôpital rule and the Taylor formula (when possible).

a) $\lim_{x \to 0} \frac{\sin x - x}{x^3}$ b) $\lim_{x \to +\infty} \frac{x - \cos x}{x + \sin^2 x}$ c) $\lim_{x \to 0} \frac{\cos x - 1 - x^2}{x^4}$ d) $\lim_{x \to 0^+} x^x$ e) $\lim_{x \to +\infty} \frac{\sqrt{x}}{x}$ f) $\lim_{x \to 0} \frac{\ln(1 + x) - x}{x^2}$ g) $\lim_{x \to 0} \arctan(x^x)$ h) $\lim_{x \to 0} \frac{x^2 \sin \frac{1}{x}}{\sin x}$ i) $\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$ j) $\lim_{x \to 0} \frac{\ln(\cos x)}{x^2}$